For some constants a and b let \[f(x) = \left\{ \begin{array}{cl} 9 - 2x & \text{if } x \le 3, \\ ax + b & \text{if } x > 3. \end{array} \right.\]The function f has the property that f(f(x)) = x for all x. What is a + b?

Respuesta :

Answer:

The value of a+b is 4.

Step-by-step explanation:

The given function is

[tex]\[f(x) = \left\{ \begin{array}{cl} 9 - 2x & \text{if } x \le 3, \\ ax + b & \text{if } x > 3. \end{array} \right.\][/tex]

It is given that for some constants a and b the function f has the property that f(f(x))=x for all x.

For x≤3,

[tex]f(x)=9-2x[/tex]

For x>3,

[tex]f(x)=ax+b[/tex]

At x=0,

[tex]f(0)=9-2(0)=9[/tex]

[tex]f(f(0))=f(9)\Rightarrow a(9)+b=9a+b[/tex]

Using property f(f(x))=x,

[tex]f(f(0))=0[/tex]

[tex]9a+b=0[/tex]                     .... (1)

At x=1,

[tex]f(1)=9-2(1)=7[/tex]

[tex]f(f(1))=f(7)\Rightarrow a(7)+b=7a+b[/tex]

Using property f(f(x))=x,

[tex]f(f(1))=1[/tex]

[tex]7a+b=1[/tex]                     .... (2)

Subtract equation (2) from equation (1).

[tex]9a+b-(7a+b)=0-1[/tex]    

[tex]2a=-1[/tex]

Divide both sides by 2.

[tex]a=-\frac{1}{2}[/tex]

Substitute this value in equation (1).

[tex]9(-\frac{1}{2})+b=0[/tex]

[tex]b=\frac{9}{2}[/tex]

The value of a is [tex]-\frac{1}{2}[/tex] and value of b is [tex]\frac{9}{2}[/tex]. The value of a+b is

[tex]a+b=-\frac{1}{2}+\frac{9}{2}[/tex]

[tex]a+b=4[/tex]

Therefore the value of a+b is 4.